Answer:
y= -3x/5
Step-by-step explanation:
y=mx+b where m is the slope and b is the y-intercept
m= rise/run = -3/5 (go from one point to the other on the graph)
y-intercept is (0,0)
Answer:
Part a) Rectangle
Part b) Triangle
Step-by-step explanation:
<u><em>The picture of the question in the attached figure N 1</em></u>
Part A) A cross section of the rectangular pyramid is cut with a plane parallel to the base. What is the name of the shape created by the cross section?
we know that
When a geometric plane slices any right pyramid so that the cut is parallel to the plane of the base, the cross section will have the same shape (but not the same size) as the base, So, in the case of a right rectangular pyramid, the cross section is a rectangle
Part b) If a cross section of the rectangular pyramid is cut perpendicular to the base, passing through the top vertex, what would be the shape of the resulting cross section?
we know that
Cross sections perpendicular to the base and through the vertex will be triangles
see the attached figure N 2 to better understand the problem
Answer:
if it's true simple interest then each year's interest is the same so interest for one year is a quarter of 241.50 which give you the rate but if each year's interest in left in the account the result will be different , it is a really poorly worded question
There is no box’s also give me brainliest answer
Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,


which is a cubic polynomial in
with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).
Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.
f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3
So, we have three subsets over which f(x) can be considered invertible.
• (-∞, -1/√3)
• (-1/√3, 1/√3)
• (1/√3, ∞)
By the inverse function theorem,

where f(a) = b.
Solve f(x) = 2 for x :
x³ - x + 2 = 2
x³ - x = 0
x (x² - 1) = 0
x (x - 1) (x + 1) = 0
x = 0 or x = 1 or x = -1
Then
can be one of
• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);
• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or
• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)